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I have a problem with this graphic, specifically the right hand side.

enter image description here

Why a "sextet" atom (an atom with 6 valence electrons)? I thought only a select few elements (C, N, and O) actually follow the octet rule.

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  • $\begingroup$ Are you referring to BF3? $\endgroup$ May 10 '14 at 5:43
  • $\begingroup$ That would be an example, yes. $\endgroup$
    – Dissenter
    May 10 '14 at 5:44
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Where you find this picture? Very misleading. When sextet is mentioned, the better way is to say is lone pair acceptor or Lewis Acid. You simply need one empty molecular orbital on the atom to be able to do all the steps in the picture. In the case for common B, C, N and O they happen to be preferring sixtet. However, for heavier element, they are wrong. If you change the $BF_{3}$ to $FeCl_{3}$ in the picture, the reaction will be also true. But the iron center is not sextet, it is $3s^{2}3p^{6}3d^{5}$ in its outer electron layer.

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  • $\begingroup$ It's the first Google result for "electron pushing" and it's from the U. Wisconsin, Madison website (not that I go there; I actually happen to be a few thousand miles away). $\endgroup$
    – Dissenter
    May 10 '14 at 14:07
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A "sextet" atom has a lack of electrons and is eager to get 2 additional valence electrons to fill its octet. By a Lewis acid-base reaction (between ether and $\ce{BF3}$) or lone-pair donation from a neighboring atom (the mesomeric structures of the two carbocations), another bond with 2 electrons to the sextet atom is made and its octet is completed.

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